Ratio, Direct & Inverse Proportion, Map Scale: O-Level/SEC E-Math

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Maths In Real Life

$A park seen from above with a winding running path and a lawn, with a map scale bar, the path length in centimetres on the map and the lawn drawn over the picture$

Maths In Real Life 1

Planning a park run

A map scale lets planners work out real distances and areas from a drawing. Lengths use the scale; areas use the square of the scale.

A park map has scale $1 : 25000$ (made-up map). The running path is $9.2$ cm long on the map and the park lawn has area $16$ cm² on the map.

(a)Find the real length of the running path in km.
(b)Find the real area of the lawn in km².
(c)A water point is $1.5$ km from the start. How far is it from the start on the map?

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Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(a)

\displaystyle \text{actual} = 9.2 \times 25000 = 230000\ \text{cm}

The scale 1 : 25000 means 1 cm on the map is 25000 cm on the ground.

\displaystyle 230000\ \text{cm} \div 100000 = 2.3\ \text{km}

Change centimetres to km: divide by 100000.

(b)

\displaystyle \text{area scale} = 1 : (25000)^2 = 1 : 625000000

The area scale is the SQUARE of the length scale.

\displaystyle \text{actual} = 16 \times 625000000 = 10000000000\ \text{cm}^2

Multiply the map area by the area scale number.

\displaystyle 10000000000 \div 10000000000 = 1\ \text{km}^2

Change cm² to km²: divide by 10000000000 (because 1 km = 100000 cm, so 1 km² = 10000000000 cm²).

(c)

\displaystyle 1.5\ \text{km} = 1.5 \times 100000 = 150000\ \text{cm}

Change km to centimetres first.

\displaystyle \text{map length} = 150000 \div 25000 = 6\ \text{cm}

Divide by the scale number to shrink the actual length onto the map.

Final answer

(a)

2.3\ \text{km}

(b)

1\ \text{km}^2

(c)

6\ \text{cm}

$Painters working along an HDB corridor with a graph of days against number of painters drawn over the picture$

Maths In Real Life 2

Painting the corridor

When a job is shared among more people, it takes fewer days. Planners use inverse proportion to schedule the work.

$5$ painters take $12$ days to paint all the corridors of a block of flats (made-up numbers). All the painters work at the same rate.

(a)How many days would $4$ painters take?
(b)The work must be finished in $6$ days. How many painters are needed?

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Colour keyadd, subtract, multiply, divide or simplify

(a)

More painters means fewer days: the number of days is inversely proportional to the number of painters.

Inverse proportion.

\displaystyle \text{total work} = 5 \times 12 = 60 \ \text{(painters} \times \text{days)}

The total amount of work does not change.

\displaystyle \text{days} = 60 \div 4 = 15

Divide the total work by the new number of painters.

(b)

The total work is the same, so painters × days is constant.

Inverse proportion.

\displaystyle 5 \times 12 = 60

Painters × days = total work.

\displaystyle 60 \div 6 = 10

Divide the total work by the number of days wanted to find the painters needed.

Final answer

(a)

15

days

(b)

10

painters

$A tray of cookies and a bag of flour on a kitchen counter with a straight line graph of flour mass against number of cookies drawn over the picture$

Maths In Real Life 3

Scaling a recipe

A recipe is a direct proportion: double the cookies and you double the flour. The recipe numbers here are made up.

A recipe uses $300$ g of flour to make $12$ cookies. The mass $m$ of flour is directly proportional to the number $n$ of cookies.

(a)How much flour is needed for $20$ cookies?
(b)How many cookies can be made with $750$ g of flour?

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Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(a)

m

is directly proportional to

n

, so

m = kn

for a constant

k

.

Direct proportion: both quantities grow (or shrink) by the same factor.

\displaystyle 300 = k \times 12 \ \Rightarrow \ k = \dfrac{ 300 }{ 12 } = 25

Use the given pair of values to find the constant k.

\displaystyle m = 25 n

Write the relationship with the value of k.

\displaystyle m = 25 \times 20 = 500

Substitute n = 20.

(b)

m

is directly proportional to

n

, so

m = kn

for a constant

k

.

Direct proportion: both quantities grow (or shrink) by the same factor.

\displaystyle 300 = k \times 12 \ \Rightarrow \ k = \dfrac{ 300 }{ 12 } = 25

Use the given pair of values to find the constant k.

\displaystyle m = 25 n

Write the relationship with the value of k.

\displaystyle 750 = 25 n \ \Rightarrow \ n = 30

Substitute m = 750 and solve for n.

Final answer

(a)

m = 500\ \text{g}

(b)

n = 30\ \text{cookies}

Supports K210-N2.1 · K210-N2.2 · K210-N2.3 · K210-N2.5 · N2.2

Ratio: equivalent ratios and simplest form

What this chapter covers (SEC G3 Mathematics, K310, topic N2)

N2.1: ratios involving rational numbers
N2.2: writing a ratio in its simplest form
N2.3: map scales (distance and area)
N2.4: direct and inverse proportion

Quick refresher (supports K310 topic N2). The ideas below are used throughout this chapter: comparison between two or more quantities by ratio; relationship between ratio and fraction; dividing a quantity in a given ratio; equivalent ratios.

A ratio compares quantities that are measured in the same unit. We write it as $a : b$ . Two ratios are equivalent when you can get one from the other by multiplying or dividing every term by the same number, like $2 : 3$ and $4 : 6$ . A ratio is in its simplest form when the terms are whole numbers with no common factor.

The Core Idea

Multiply or divide every term by the same number and the ratio does not change.

Divide by the HCF of all the terms to reach the simplest form.

Example 1.1

Simplest form

Write each ratio in its simplest form.

(i) $48 : 36$
(ii) $24 : 36 : 60$

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Colour keyfactorise, HCF or LCM

(i)

\displaystyle \text{HCF of } 48, 36 = 12

The HCF is the biggest number that divides every term: use prime factors from Chapter 1.

\displaystyle 48 : 36 = 48 \div 12 : 36 \div 12 = 4 : 3

Divide every term by the HCF. The ratio keeps its meaning but uses the smallest whole numbers.

(ii)

\displaystyle \text{HCF of } 24, 36, 60 = 12

The HCF is the biggest number that divides every term: use prime factors from Chapter 1.

\displaystyle 24 : 36 : 60 = 24 \div 12 : 36 \div 12 : 60 \div 12 = 2 : 3 : 5

Divide every term by the HCF. The ratio keeps its meaning but uses the smallest whole numbers.

Final answer

(i)

4 : 3

(ii)

2 : 3 : 5

Your Turn 1.1

Write each ratio in its simplest form.

(i) $54 : 72$
(ii) $28 : 42 : 70$

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Colour keyfactorise, HCF or LCM

(i)

\displaystyle \text{HCF of } 54, 72 = 18

The HCF is the biggest number that divides every term: use prime factors from Chapter 1.

\displaystyle 54 : 72 = 54 \div 18 : 72 \div 18 = 3 : 4

Divide every term by the HCF. The ratio keeps its meaning but uses the smallest whole numbers.

(ii)

\displaystyle \text{HCF of } 28, 42, 70 = 14

The HCF is the biggest number that divides every term: use prime factors from Chapter 1.

\displaystyle 28 : 42 : 70 = 28 \div 14 : 42 \div 14 : 70 \div 14 = 2 : 3 : 5

Divide every term by the HCF. The ratio keeps its meaning but uses the smallest whole numbers.

Final answer

(i)

3 : 4

(ii)

2 : 3 : 5

Example 1.2

Finding a missing term

Find the value of $x$ in each pair of equivalent ratios.

(i) $3 : 5 = x : 35$
(ii) $4 : 7 = 20 : x$

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Colour keyadd, subtract, multiply, divide or simplifyround, estimate or compare

(i)

\displaystyle 35 \div 5 = 7

Compare the known terms in the same position: the scale factor is 7.

\displaystyle x = 3 \times 7 = 21

Multiply the other term by the same scale factor to keep the ratio equivalent.

(ii)

\displaystyle 20 \div 4 = 5

Compare the known terms in the same position: the scale factor is 5.

\displaystyle x = 7 \times 5 = 35

Multiply the other term by the same scale factor to keep the ratio equivalent.

Final answer

(i)

x = 21

(ii)

x = 35

Ratio and fraction

In the ratio $3 : 5$ there are $3 + 5 = 8$ parts altogether, so the first share is $\frac{ 3 }{ 8 }$ of the whole, not $\frac{ 3 }{ 5 }$ .

Think About It

Is the ratio

3 : 5

the same as the fraction

\frac{ 3 }{ 5 }

?

Show the worked answerHide the worked answer

Colour keyadd, subtract, multiply, divide or simplify

\displaystyle 3 + 5 = 8

Add the parts to get the total number of parts.

\displaystyle \text{first share} = \dfrac{ 3 }{ 8 }

A share is its parts out of ALL the parts, not out of the other share.

Final answerNo. The ratio has 8 parts in all, so the first share is a fraction with 8 on the bottom.

Example 1.3

Ratio as a fraction of the whole

Ali and Ben share some sweets in the ratio $3 : 5$ .

(i)What fraction of the sweets does Ali get?
(ii)There are $64$ sweets. How many does Ben get?

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Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle 3 + 5 = 8

Add the parts to get the total number of parts.

\displaystyle \text{the first share} = \dfrac{ 3 }{ 8 }, \quad \text{the second share} = \dfrac{ 5 }{ 8 }

Each part is that many parts out of the total number of parts.

(ii)

\displaystyle 3 + 5 = 8

Total number of parts.

\displaystyle \dfrac{ 5 }{ 8 } \times 64 = 40

Fraction of the total × the total.

Final answer

(i)The first share is

\frac{ 3 }{ 8 }

of the total and the second is

\frac{ 5 }{ 8 }

of the total.

(ii)

40

Your Turn 1.3

Mei and Raj share some stickers in the ratio $2 : 7$ .

(i)What fraction of the stickers does Mei get?
(ii)There are $72$ stickers. How many does Raj get?

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Colour keyadd, subtract, multiply, divide or simplify

(i)

\displaystyle 2 + 7 = 9

Add the parts to get the total number of parts.

\displaystyle \text{the first share} = \dfrac{ 2 }{ 9 }, \quad \text{the second share} = \dfrac{ 7 }{ 9 }

Each part is that many parts out of the total number of parts.

(ii)

\displaystyle 2 + 7 = 9

Total number of parts.

\displaystyle \dfrac{ 7 }{ 9 } \times 72 = 56

Fraction of the total × the total.

Final answer

(i)The first share is

\frac{ 2 }{ 9 }

of the total and the second is

\frac{ 7 }{ 9 }

of the total.

(ii)

56

Links To

Want the short version first? Read the matching part of the free E-Math revision page: numbers and arithmetic

Supports N2.1 · N2.2 · K210-N2.5

Ratios of rational numbers

Ratios often involve fractions, decimals or different units. Turn them into whole numbers first: multiply by a number that clears the fractions or decimal points, convert to a common unit, then simplify. When two ratios share a quantity, match that quantity to combine them.

The Core Idea

Clear the fractions or decimals, make the units the same, then simplify.

The ratio of two rational numbers is a ratio of whole numbers once you scale up.

Example 2.1

A ratio of fractions

Write $\frac{ 2 }{ 3 } : \frac{ 5 }{ 6 }$ as a ratio of whole numbers in simplest form.

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Colour keyfactorise, HCF or LCM

\displaystyle \frac{2}{3} : \frac{5}{6} = \frac{2}{3} \times 6 : \frac{5}{6} \times 6

Multiply every term by the LCM of the denominators, 6, to clear the fractions.

\displaystyle = 4 : 5

All the terms are now whole numbers.

Final answer

4 : 5

Example 2.2

A ratio of decimals

Write $0.45 : 0.6$ in simplest form.

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Colour keyadd, subtract, multiply, divide or simplifyfactorise, HCF or LCM

\displaystyle 0.45 : 0.6 = 0.45 \times 20 : 0.6 \times 20

Multiply every term by 20 (a number that makes every decimal a whole number) to clear the decimal points.

\displaystyle = 9 : 12

All the terms are now whole numbers.

\displaystyle = 3 : 4

Divide by the HCF, 3, to reach the simplest form.

Final answer

3 : 4

Example 2.3

Different units

A film lasts $1.5$ hours and an advert break lasts $40$ minutes. Find the ratio of the film time to the break time in simplest form.

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Colour keyfactorise, HCF or LCMexpand, rewrite, substitute or solve

\displaystyle 1.5\ \text{hours} = 1.5 \times 60 = 90\ \text{minutes}

There are 60 minutes in 1 hour: change hours to minutes so both are in the same unit.

\displaystyle 90 : 40

Now both quantities are in the same unit, so the units can be dropped.

\displaystyle \text{HCF of } 90, 40 = 10

The HCF is the biggest number that divides every term: use prime factors from Chapter 1.

\displaystyle 90 : 40 = 90 \div 10 : 40 \div 10 = 9 : 4

Divide every term by the HCF. The ratio keeps its meaning but uses the smallest whole numbers.

Final answer

9 : 4

Your Turn 2.3

A walk takes $2.5$ km and a jog takes $800$ m. Find the ratio of the walk distance to the jog distance in simplest form.

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Colour keyfactorise, HCF or LCMexpand, rewrite, substitute or solve

\displaystyle 2.5\ \text{km} = 2.5 \times 1000 = 2500\ \text{m}

There are 1000 metres in 1 kilometre: change km to m so both are in the same unit.

\displaystyle 2500 : 800

Now both quantities are in the same unit, so the units can be dropped.

\displaystyle \text{HCF of } 2500, 800 = 100

The HCF is the biggest number that divides every term: use prime factors from Chapter 1.

\displaystyle 2500 : 800 = 2500 \div 100 : 800 \div 100 = 25 : 8

Divide every term by the HCF. The ratio keeps its meaning but uses the smallest whole numbers.

Final answer

25 : 8

Example 2.4

Joining two ratios

$a : b = 3 : 4$ and $b : c = 6 : 7$ . Find $a : b : c$ .

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Colour keyfactorise, HCF or LCMexpand, rewrite, substitute or solve

\displaystyle \text{LCM of } 4 \text{ and } 6 = 12

Make the shared quantity (the middle term) the same in both ratios.

\displaystyle 3 : 4 = 9 : 12 \quad (\times 3), \qquad 6 : 7 = 12 : 14 \quad (\times 2)

Scale each ratio so that its middle term equals the LCM.

\displaystyle 9 : 12 : 14

Now the middle terms match, so the two ratios can be joined.

Final answer

9 : 12 : 14

SGSchoolKaki Tip

Always check the answer by reading it back: does the first ratio still appear when you take the first two terms?

Supports N2.1 · K210-N2.3

Sharing in a ratio and ratio problems

To share a quantity in a ratio, add the parts, find the value of ONE part, then multiply. Many ratio problems give a difference, or a change to one quantity, instead of a total. In every case, find what one part is worth.

The Core Idea

Total ÷ total parts = one part. Each share = its parts × one part.

If you are given a difference instead, divide it by the number of parts it covers.

Example 3.1

Sharing money in a ratio

Mei and Raj share $\text{\textdollar}180.00$ in the ratio $3 : 5$ . Find how much each person gets.

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Colour keyadd, subtract, multiply, divide or simplify

\displaystyle 3 + 5 = 8

Add the parts to find the total number of parts.

\displaystyle \text{1 part} = \text{\textdollar}180.00 \div 8 = \text{\textdollar}22.50

Divide the total by the number of parts to find the value of one part.

\displaystyle { 3 \times \text{\textdollar}22.50 = \text{\textdollar}67.50 }, \allowbreak \quad { 5 \times \text{\textdollar}22.50 = \text{\textdollar}112.50 }

Multiply by the number of parts in each share.

\displaystyle \text{\textdollar}67.50 + \text{\textdollar}112.50 = \text{\textdollar}180.00

Check: the shares add up to the total.

Final answer

Mei:

\text{\textdollar}67.50

; Raj:

\text{\textdollar}112.50

Your Turn 3.1

Anna and Ben share $\text{\textdollar}252.00$ in the ratio $4 : 5$ . Find how much each person gets.

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplify

\displaystyle 4 + 5 = 9

Add the parts to find the total number of parts.

\displaystyle \text{1 part} = \text{\textdollar}252.00 \div 9 = \text{\textdollar}28.00

Divide the total by the number of parts to find the value of one part.

\displaystyle { 4 \times \text{\textdollar}28.00 = \text{\textdollar}112.00 }, \allowbreak \quad { 5 \times \text{\textdollar}28.00 = \text{\textdollar}140.00 }

Multiply by the number of parts in each share.

\displaystyle \text{\textdollar}112.00 + \text{\textdollar}140.00 = \text{\textdollar}252.00

Check: the shares add up to the total.

Final answer

Anna:

\text{\textdollar}112.00

; Ben:

\text{\textdollar}140.00

Example 3.2

A three-way share

$96$ kg of rice is shared among A, B and C in the ratio $2 : 3 : 5$ . Find each share.

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplify

\displaystyle 2 + 3 + 5 = 10

Add the parts to find the total number of parts.

\displaystyle \text{1 part} = 96\ \text{kg} \div 10 = 9.6\ \text{kg}

Divide the total by the number of parts to find the value of one part.

\displaystyle { 2 \times 9.6 = 19.2 }, \allowbreak \quad { 3 \times 9.6 = 28.8 }, \allowbreak \quad { 5 \times 9.6 = 48 }

Multiply by the number of parts in each share.

\displaystyle 19.2 + 28.8 + 48 = 96

Check: the shares add up to the total.

Final answer

A:

19.2\ \text{kg}

; B:

28.8\ \text{kg}

; C:

48\ \text{kg}

Your Turn 3.2

$120$ kg of rice is shared among A, B and C in the ratio $1 : 4 : 7$ . Find each share.

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplify

\displaystyle 1 + 4 + 7 = 12

Add the parts to find the total number of parts.

\displaystyle \text{1 part} = 120\ \text{kg} \div 12 = 10\ \text{kg}

Divide the total by the number of parts to find the value of one part.

\displaystyle { 1 \times 10 = 10 }, \allowbreak \quad { 4 \times 10 = 40 }, \allowbreak \quad { 7 \times 10 = 70 }

Multiply by the number of parts in each share.

\displaystyle 10 + 40 + 70 = 120

Check: the shares add up to the total.

Final answer

A:

10\ \text{kg}

; B:

40\ \text{kg}

; C:

70\ \text{kg}

Example 3.3

A difference is given

Two friends save in the ratio $2 : 5$ . The second has $\text{\textdollar}45.00$ more than the first. Find the total saved.

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Colour keyadd, subtract, multiply, divide or simplify

\displaystyle 5 - 2 = 3

The difference between the two shares is this many parts.

\displaystyle 3 \text{ part(s)} = \text{\textdollar}45.00 \ \Rightarrow \ \text{1 part} = \text{\textdollar}15.00

Divide the known difference by the number of parts it covers.

\displaystyle \text{total} = 7 \times \text{\textdollar}15.00 = \text{\textdollar}105.00

Multiply by the total number of parts.

Final answer

The total is

\text{\textdollar}105.00

.

Your Turn 3.3

Two friends save in the ratio $3 : 7$ . The second has $\text{\textdollar}64.00$ more than the first. Find the total saved.

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplify

\displaystyle 7 - 3 = 4

The difference between the two shares is this many parts.

\displaystyle 4 \text{ part(s)} = \text{\textdollar}64.00 \ \Rightarrow \ \text{1 part} = \text{\textdollar}16.00

Divide the known difference by the number of parts it covers.

\displaystyle \text{total} = 10 \times \text{\textdollar}16.00 = \text{\textdollar}160.00

Multiply by the total number of parts.

Final answer

The total is

\text{\textdollar}160.00

.

Example 3.4

A quantity changes

A bag has red and blue marbles in the ratio $7 : 4$ . After $12$ blue marbles are added, the ratio becomes $7 : 6$ . How many marbles of each colour were there at first?

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Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

\displaystyle \text{red} : \text{blue} = { 7 : 4 \ \text{before}}, \allowbreak \qquad { 7 : 6 \ \text{after}}

Write the ratio before and after.

The number of red does not change, and its parts are already equal in both ratios.

Match the unchanged quantity.

\displaystyle \text{after} = 7 : 6

The unchanged part is now the same in both ratios.

\displaystyle 6 - 4 = 2 \ \text{parts}

This many parts are the 12 that were added.

\displaystyle 2 \text{ part(s)} = 12 \ \Rightarrow \ \text{1 part} = 6

Find the value of one part.

\displaystyle {\text{red}: 7 \times 6 = 42 }, \allowbreak \qquad {\text{blue}: 4 \times 6 = 24 }

Original amounts = parts × value of one part.

Final answer

Originally: red

42

, blue

24

.

Your Turn 3.4

A box has green and yellow beads in the ratio $3 : 4$ . After $10$ yellow beads are added, the ratio becomes $3 : 5$ . How many beads of each colour were there at first?

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Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

\displaystyle \text{green} : \text{yellow} = { 3 : 4 \ \text{before}}, \allowbreak \qquad { 3 : 5 \ \text{after}}

Write the ratio before and after.

The number of green does not change, and its parts are already equal in both ratios.

Match the unchanged quantity.

\displaystyle \text{after} = 3 : 5

The unchanged part is now the same in both ratios.

\displaystyle 5 - 4 = 1 \ \text{parts}

This many parts are the 10 that were added.

\displaystyle 1 \text{ part(s)} = 10 \ \Rightarrow \ \text{1 part} = 10

Find the value of one part.

\displaystyle {\text{green}: 3 \times 10 = 30 }, \allowbreak \qquad {\text{yellow}: 4 \times 10 = 40 }

Original amounts = parts × value of one part.

Final answer

Originally: green

30

, yellow

40

.

Example 3.5

The unchanged part needs matching

A class has boys and girls in the ratio $4 : 5$ . After $6$ girls join, the ratio becomes $2 : 3$ . How many boys and girls were there at first?

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

\displaystyle \text{boys} : \text{girls} = { 4 : 5 \ \text{before}}, \allowbreak \qquad { 2 : 3 \ \text{after}}

Write the ratio before and after.

The number of boys does not change, so make its parts equal in both ratios: multiply the new ratio by

2

.

Match the unchanged quantity.

\displaystyle \text{after} = 4 : 6

The unchanged part is now the same in both ratios.

\displaystyle 6 - 5 = 1 \ \text{parts}

This many parts are the 6 that were added.

\displaystyle 1 \text{ part(s)} = 6 \ \Rightarrow \ \text{1 part} = 6

Find the value of one part.

\displaystyle {\text{boys}: 4 \times 6 = 24 }, \allowbreak \qquad {\text{girls}: 5 \times 6 = 30 }

Original amounts = parts × value of one part.

Final answer

Originally: boys

24

, girls

30

.

Your Turn 3.5

A club has men and women in the ratio $3 : 5$ . After $10$ women join, the ratio becomes $1 : 2$ . How many men and women were there at first?

Show worked solutionHide worked solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

\displaystyle \text{men} : \text{women} = { 3 : 5 \ \text{before}}, \allowbreak \qquad { 1 : 2 \ \text{after}}

Write the ratio before and after.

The number of men does not change, so make its parts equal in both ratios: multiply the new ratio by

3

.

Match the unchanged quantity.

\displaystyle \text{after} = 3 : 6

The unchanged part is now the same in both ratios.

\displaystyle 6 - 5 = 1 \ \text{parts}

This many parts are the 10 that were added.

\displaystyle 1 \text{ part(s)} = 10 \ \Rightarrow \ \text{1 part} = 10

Find the value of one part.

\displaystyle {\text{men}: 3 \times 10 = 30 }, \allowbreak \qquad {\text{women}: 5 \times 10 = 50 }

Original amounts = parts × value of one part.

Final answer

Originally: men

30

, women

50

.

SGSchoolKaki Tip

When one quantity changes, do not add the change to the ratio terms. The change is a number of items, so it must be matched with the right number of parts.

Supports N2.3

Map scales: distance

A map scale links a length on the map to the real length on the ground. The ratio scale $1 : n$ means that 1 cm on the map stands for $n$ cm on the ground. Convert centimetres to metres or kilometres at the end: 100 cm = 1 m and 100 000 cm = 1 km.

The Core Idea

Map length × n = real length. Real length ÷ n = map length.

Work in centimetres, then convert to metres or kilometres.

Figure 1: a park drawn at scale 1 : 50 000. The scale bar shows how far 4 cm on the map is on the ground.

Example 4.1

From the map to the ground

The scale of a map is $1 : 50000$ . Two MRT stations are $6.4$ cm apart on the map. Find the real distance in km.

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplify

\displaystyle \text{actual} = 6.4 \times 50000 = 320000\ \text{cm}

The scale 1 : 50000 means 1 cm on the map is 50000 cm on the ground.

\displaystyle 320000\ \text{cm} \div 100000 = 3.2\ \text{km}

Change centimetres to km: divide by 100000.

Final answer

3.2\ \text{km}

Example 4.2

From the ground to the map

The scale of a map is $1 : 50000$ . A road is $4.5$ km long. How long is it on the map, in cm?

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

\displaystyle 4.5\ \text{km} = 4.5 \times 100000 = 450000\ \text{cm}

Change km to centimetres first.

\displaystyle \text{map length} = 450000 \div 50000 = 9\ \text{cm}

Divide by the scale number to shrink the actual length onto the map.

Final answer

9\ \text{cm}

Example 4.3

Finding the scale

On a map, $3$ cm represents $1.2$ km. Write the scale in the form $1 : n$ .

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

\displaystyle 1.2\ \text{km} = 120000\ \text{cm}

Change km to centimetres so both lengths use the same unit.

\displaystyle 3\ \text{cm} : 120000\ \text{cm} = 1 : 120000 \div 3

Divide both terms by the map length to get 1 on the left.

\displaystyle = 1 : 40000

This is the ratio scale.

Final answer

1 : 40000

Example 4.4

Two questions on one map

A running track measures $36$ cm on a map with scale $1 : 20000$ .

(i)Find the real length of the track in km.
(ii)A runner covers $5$ km. How long would that be on the map?

Show solutionHide solution

Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle \text{actual} = 36 \times 20000 = 720000\ \text{cm}

The scale 1 : 20000 means 1 cm on the map is 20000 cm on the ground.

\displaystyle 720000\ \text{cm} \div 100000 = 7.2\ \text{km}

Change centimetres to km: divide by 100000.

(ii)

\displaystyle 5\ \text{km} = 5 \times 100000 = 500000\ \text{cm}

Change km to centimetres first.

\displaystyle \text{map length} = 500000 \div 20000 = 25\ \text{cm}

Divide by the scale number to shrink the actual length onto the map.

Final answer

(i)

7.2\ \text{km}

(ii)

25\ \text{cm}

Your Turn 4.4

A cycling path measures $28$ cm on a map with scale $1 : 40000$ .

(i)Find the real length of the path in km.
(ii)A cyclist covers $6$ km. How long would that be on the map?

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Colour keyadd, subtract, multiply, divide or simplifyexpand, rewrite, substitute or solve

(i)

\displaystyle \text{actual} = 28 \times 40000 = 1120000\ \text{cm}

The scale 1 : 40000 means 1 cm on the map is 40000 cm on the ground.

\displaystyle 1120000\ \text{cm} \div 100000 = 11.2\ \text{km}

Change centimetres to km: divide by 100000.

(ii)

\displaystyle 6\ \text{km} = 6 \times 100000 = 600000\ \text{cm}

Change km to centimetres first.

\displaystyle \text{map length} = 600000 \div 40000 = 15\ \text{cm}

Divide by the scale number to shrink the actual length onto the map.

Final answer

(i)

11.2\ \text{km}

(ii)

15\ \text{cm}

SGSchoolKaki Tip

Keep the units straight: the scale 1 : 50 000 compares centimetres with centimetres. Convert to km only at the end.

Ratio and Proportion: O-Level E-Math Notes

Maths In Real Life

Planning a park run

Painting the corridor

Scaling a recipe

Ratio: equivalent ratios and simplest form

What this chapter covers (SEC G3 Mathematics, K310, topic N2)

Simplest form

Finding a missing term

Ratio and fraction

Ratio as a fraction of the whole

Ratios of rational numbers

A ratio of fractions

A ratio of decimals

Different units

Joining two ratios

Map scales: distance

From the map to the ground

From the ground to the map

Finding the scale

Two questions on one map

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